66. Using the bond energies in Table 7.3, determine the approximate enthalpy change for each of the following reactions: (a) Cl, (g) + 3F,(g) → 2CIF,(g) (b) H,C= CH,(g) + H,(g) - — Н,ССН, (g) Solution (a) -807 kJ; DH° = ED De-c + 4Dc-H + Dµ-H - Dc-c - 6Dc-H : 611 + 4(415) + 436 – 345 – 6(415) = -128 kJ bonds broken bonds formed (b) %3D
66. Using the bond energies in Table 7.3, determine the approximate enthalpy change for each of the following reactions: (a) Cl, (g) + 3F,(g) → 2CIF,(g) (b) H,C= CH,(g) + H,(g) - — Н,ССН, (g) Solution (a) -807 kJ; DH° = ED De-c + 4Dc-H + Dµ-H - Dc-c - 6Dc-H : 611 + 4(415) + 436 – 345 – 6(415) = -128 kJ bonds broken bonds formed (b) %3D
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 69E: Using the data in Appendix G, calculate the standard enthalpy change for each of the following...
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Question
How did we get 807kJ from question a?how did we get 128kJ for b ?is there a step by step explanation?
![66. Using the bond energies in Table 7.3, determine the approximate enthalpy change for each of
the following reactions:
(a) Cl, (g) + 3F,(g)
→ 2CIF, (g)
(b) H,C=CH,(g) + H,(g) → H,CCH,(g)
Solution
(a) -807 kJ;
ED
De-c + 4Dc-H + Du-H - Dc-c - 6Dc-H:
= 611 + 4(415) + 436 -
= -128 kJ
DH°
ZDbonds formed
bonds broken
%3D
C-C
(b)
345 - 6(415)
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F74dd441d-ba90-4318-913b-2724bd118648%2F4pwonba_processed.jpeg&w=3840&q=75)
Transcribed Image Text:66. Using the bond energies in Table 7.3, determine the approximate enthalpy change for each of
the following reactions:
(a) Cl, (g) + 3F,(g)
→ 2CIF, (g)
(b) H,C=CH,(g) + H,(g) → H,CCH,(g)
Solution
(a) -807 kJ;
ED
De-c + 4Dc-H + Du-H - Dc-c - 6Dc-H:
= 611 + 4(415) + 436 -
= -128 kJ
DH°
ZDbonds formed
bonds broken
%3D
C-C
(b)
345 - 6(415)
|
![Bond
Bond Energy
Bond
Bond Energy
Bond
Bond Energy
H-H
436
C-S
260
F-CI
255
H-C
415
C-CI
330
F-Br
235
H-N
390
C-Br
275
Si-Si
230
H-O
464
C-I
240
SI-P
215
H-F
569
N-N
160
Si-S
225
H-Si
395
N = N
418
Si-CI
359
H-P
320
N = N
946
Si-Br
290
H-S
340
N-O
200
Si-l
215
H-CI
432
N-F
270
P-P
215
H-Br
370
NAP
210
P-S
230
H-I
295
N-CI
200
P-CI
330
C-C
345
N-Br
245
P-Br
270
C = C
611
O-0
140
P-I
215
C = C
O = 0
S-S
215
837
498
C-N
290
O-F
160
S-CI
250
C = N
615
O-Si
370
S-Br
215
C = N
O-P
350
CI-CI
243
891
C-O
350
O-CI
205
CI-Br
220
C = 0
741
200
CI-I
210
MacBook Air
C = 0
1080
F-F
160
Br-Br
190
C-F
439
F-Si
540
Br-l
180
C-Si
360
F-P
489
150
C-P
265
F-S
285](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2F74dd441d-ba90-4318-913b-2724bd118648%2Fc4r9uo6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Bond
Bond Energy
Bond
Bond Energy
Bond
Bond Energy
H-H
436
C-S
260
F-CI
255
H-C
415
C-CI
330
F-Br
235
H-N
390
C-Br
275
Si-Si
230
H-O
464
C-I
240
SI-P
215
H-F
569
N-N
160
Si-S
225
H-Si
395
N = N
418
Si-CI
359
H-P
320
N = N
946
Si-Br
290
H-S
340
N-O
200
Si-l
215
H-CI
432
N-F
270
P-P
215
H-Br
370
NAP
210
P-S
230
H-I
295
N-CI
200
P-CI
330
C-C
345
N-Br
245
P-Br
270
C = C
611
O-0
140
P-I
215
C = C
O = 0
S-S
215
837
498
C-N
290
O-F
160
S-CI
250
C = N
615
O-Si
370
S-Br
215
C = N
O-P
350
CI-CI
243
891
C-O
350
O-CI
205
CI-Br
220
C = 0
741
200
CI-I
210
MacBook Air
C = 0
1080
F-F
160
Br-Br
190
C-F
439
F-Si
540
Br-l
180
C-Si
360
F-P
489
150
C-P
265
F-S
285
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